73. Set Matrix Zeroes

矩阵置零算法

weixin_34050005

于 2019-02-12 16:57:42 发布

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文章标签： java runtime 开发工具

原文链接：https://segmentfault.com/a/1190000018132340

本文介绍了一种在原地修改矩阵的算法，当矩阵中的某个元素为0时，将其所在行和列的所有元素设置为0。文章提供了一种使用常量空间的解决方案，详细解释了如何利用矩阵的第一行和第一列作为标记，避免使用额外的空间。

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

Input: 
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output: 
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]

Example 2:

Input: 
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output: 
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]

Follow up:

A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

难度：medium

题目：给定m * n的矩阵，如果某一元素为0，则将其所在行及列都设为0。在原矩阵上执行。

一种直接的解法是空间复杂度为O(mn)
一种简单的改进是空间复杂度为O(m + n),但是仍不是最好的。
是否可以用常量空间给出解法。

思路：先统计首行，首列是否含有0。然后用首行首列来记录其它行列。

Runtime: 1 ms, faster than 99.98% of Java online submissions for Set Matrix Zeroes.
Memory Usage: 45.6 MB, less than 0.96% of Java online submissions for Set Matrix Zeroes.

class Solution {
    public void setZeroes(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int row0 = 1, column0 = 1;
        // first row
        for (int i = 0; i < n; i++) {
            if (0 == matrix[0][i]) {
                row0 = 0;
            }
        }
        // first column
        for (int i = 0; i < m; i++) {
            if (0 == matrix[i][0]) {
                column0 = 0;
            }
        }
        // other rows and columns
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (0 == matrix[i][j]) {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }
        // set 0 for other rows and columns
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (0 == matrix[0][j] || 0 == matrix[i][0]) {
                    matrix[i][j] = 0;
                }
            }
        }
        // set 0 for first row
        for (int i = 0; 0 == row0 && i < n; i++) {
            matrix[0][i] = 0;
        }
        // set 0 for first column
        for (int i = 0; 0 == column0 && i < m; i++) {
            matrix[i][0] = 0;
        }
    }
}